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Mathematics Revision Guides – Solving Trigonometric Equations of 17

Author: Mark Kudlowski

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Mathematics Revision Guides

Level: AS / A Level

AQA : C2 Edexcel: C2 OCR: C2 OCR MEI: C2

SOLVING TRIGONOMETRIC EQUATIONS

Version : 3.2 Date: 25-03-2013

Examples 4, 6, 8, 9 and 10 are copyrighted to their respective owners and used with their permission.



Trigonometric identities.

There are two very important trigonometric identities.

For all angles

cos2+ sin

2= 1 (This is the Pythagorean identity)

The square on the hypotenuse in the triangle above is 1.

The sum of the squares on the other two sides = cos2+ sin

2

The length of the opposite side to A = sin and the length of the adjacent to A = cos .

Since the tangent is the opposite divided by the adjacent, it also follows that

tan sin

cos

These two identities are important when simplifying expressions or solving various types of equations.



SOLVING TRIGONOMETRIC EQUATIONS

Any trigonometric equation of the form sin x° = a (where -1 acos x° = a (where -1 a

or tan x° = a can have an infinite variety of solutions.

To solve such equations using a calculator, it must be remembered that only one of an infinite number

of possible values will be given – we must find all solutions within the range of the question.

The graph above shows several of the possible solutions of sin x° = 0.5. Apart from the one of 30°

given on the calculator (the principal value), the possible solutions also include (180-30)° or 150°

because of the reflective symmetry of the graph about the line x = 90°. Moreover, as the graph has a

repeating period of 360°, other solutions are 390°, 510° and so on in the positive x-direction, and –

210°, -330° and so forth in the negative x-direction.

We can also use CAST

diagrams to show the two

solutions in the range 0° -

360°, and we can add or

subtract multiples of 360° as

desired for any further values,

should the question ask for

them.

Positive angles are measured

anticlockwise from the origin;

negative ones clockwise.

In general, if sin x° = a, then :

sin (180-x)° = a

sin (180n + x)° = a for even integer n ; sin (180n - x)° = a for odd integer n.

Therefore, for instance, sin (360+x)°, sin (540-x)° and sin (-180-x)° are also equal to sin x.

In radians, the solutions are given as /6 and ( – /6) or 5/6 . The graph repeats every 2radians,

and other solutions are 13/6, 17/6 (positive x-direction) and -7/6, -11/6 (negative x-direction).

In general, if sin x = a, then :

sin (-x) = a

sin (n+ x) = a for even integer n ; sin (n - x) = a for odd integer n.

Therefore, for instance, sin (2x) sin (3-x) and sin (--x) are also equal to sin x.



x°180 360 540 720

y

-1

-0.5

0.5

1

Line of symmetry: x = 0°

60°

y = cos x °

y =0.5

x = 180° x = 360° x = 720°

300° 420° 660°

x = 540°

The cosine function also shows similar periodicity.

Several of the possible solutions of cos x° = 0.5 are shown in the graph above. Apart from the one of

60° given on the calculator(the principal value), the possible solutions also include 300° because the

graph has reflective symmetry about the line x = 180°. Again, there is a repeating period of 360°, and

so other solutions are 420°, 660° and so on in the positive x-direction, and -60 ,–420- ,°300ۥ° and so

forth in the negative x-direction.

The CAST diagrams to

show the two solutions in

the range 0° - 360°.

Again, we can add or

subtract multiples of 360°

if required.

The relationship between

cos 60° and cos -60° is

evident from the left-hand

diagram.

In general, if cos x° = a, then :

cos (-x)° = a or cos (360-x)° = a

cos (360n + x)° = a for any integer n

cos (360n - x)° = a for any integer n.

Therefore, for instance, cos (360+x)°, cos (720-x)° and cos (-360-x)° are also equal to cos x.

In radians, the solutions are given as /3 and –/3 (or 5/3). The graph repeats every 2radians, and

other solutions are 5/3, 7/3 (positive x-direction) and -5/3, -7/3 (negative x-direction).

In general, if cos x = a, then :

cos (-x) = a or cos (2-x)° = a

cos (2n+ x) = a for any integer n

cos (2n - x) = a for any integer n.

Therefore, for instance, cos (2+x), cos (4x) and cos (-2-x) are also equal to cos x.



The symmetrical pairing of solutions in the range 0° x (or 0 xin radians) is common to

both the sine and cosine graphs, for all angles on either side of the graphs’ lines of symmetry.

(An exception occurs when sin x and cos x = ±1 ; then there is only one solution.in that range.)

Thus sin 115° = sin 65° (both equidistant from line of symmetry at 90°), and cos 155° = cos 205°

(both equidistant from line of symmetry at 180°).

The tangent function, on the other hand, has a period of 180° rather than 360° as in the sine and cosine

functions. Also, there is no linear symmetry, so there is no ‘doubling-up’ of solutions (as in sine and

cosine examples).

The graph shows several solutions of the equation tan x° = 1.

Apart from the one of 45° given on the calculator, there are other possible solutions because the graph

has a repeating period of 180°. Those solutions are 225°, 405° and so on in the positive x-direction, and

–135°, -315° and so forth in the negative x-direction.

The CAST diagrams for the tangent

function are simpler in form than the others

– you merely add multiples of 180° to

obtain all the solutions needed.

In general, if tan x° = a, then tan (180n + x)° = a for any integer n.

In radians, the calculator solution is /4 . The graph repeats every radians, and other solutions are

5/4, 9/4 (positive x-direction) and -3/4, -7/4 (negative x-direction).

In general, if tan x = a, then tan (n + x) = a for any integer n.



Example (1): Solve sin x° = 0.85 for 0° x°, giving answers in degrees to one decimal place.

The principal value is 58.2°, but care must be taken to ensure that all solutions within the range are

given.

The substitution formulae can be used, but a sketch graph or CAST diagram will probably be easier in

finding extra solutions.

The two important lines of symmetry here are x = 90° and x = 450°.

Since 58.2° is 31.8° less than 90°, the solution on the other side of that line of symmetry must be 31.8°

greater than 90°, or 121.8°.

Alternatively we can use sin (180-x)° = sin x°, and 180° - 58.2° = 121.8°.

CAST diagram illustration:

Having obtained the two solutions above, it is a simple matter of adding and subtracting multiples of

360° as required. Subtracting 360° is no help as there will be no new values found within the range,

but adding 360° will give two other solutions:

(58.2 + 360)° or 418.2°.

(121.8 + 360)° or 481.8°.

the solutions of sin x° = 0.85 for 0° x0° are 58.2°, 121.8°, 418.2° and 481.8°.



Example (2): Solve sin x° = -0.8 for 0° x360°, giving answer in degrees to one decimal place.

The principal value, and the one given on a calculator, is -53.1°, from which we can derive the other

solutions. Note that this solution is not in the quoted range, and so we must add an appropriate multiple

of 360° (the period of sin x) to it.

Here, adding 360° gives one solution, i.e. 306.9°.

x°-60 60 120 180 240 300 360

y

-1

-0.5

0.5

1y = sin x °

y = -0.8233.1° 306.9°

Line of symmetry: x = 270°

The important line of symmetry here is x = 270°.

Since 306.9° is 36.9° greater than 270°, the solution on the other side of that line of symmetry must be

36.9° less than 270°, or 233.1°.

the solutions of sin x° = -0.8 for 0° x360° are 233.1° and 306.9°.

Again we could have used sin (180-x)° = sin x°, and 180° - (-53.1°) = 233.1°, although it is less

obvious from the diagram.




Example (3): Solve cos x° = -0.66 for 0° x720°, giving answer in degrees to one decimal place.

The two important lines of symmetry here are x = 180° and x = 540°.

Here, the principal value, and the one given on a calculator, is 131.3°.

Since 131.3° is 48.7° less than 180°, the solution on the other side of that line of symmetry must be

48.7° greater than 180°, or 228.7°.

We could have used cos (360-x)° = cos x°, and 360° - 131.3° = 228.7°.

Having obtained the two solutions above, it is a simple matter of adding and subtracting multiples of

360° as required. Subtracting 360° is no help as there will be no new values found within the range,

but adding 360° will give two other solutions, i.e. 491.3° and 588.7°.

the solutions of cos x° = -0.66 for 0° x0° are 131.3°, 228.7°, 491.3° and 588.7°.




Example (4): Solve cos x = 0.75 for 0 x, giving answers in radians to 3 decimal places.

The principal value is 0.723 c , but to find the other solution, we can make use of the lines of symmetry

at x = 0 and x = 2Since the principal value is 0.723 greater than zero, the other required value

must be 0.723 less than 2or 5.560c. (This is the same as using the identity cos (2-x) = cos x.)

We could have also used the line of symmetry at x = and worked the value as + ( -0.723) ,

leading to the same conclusion.

(This, and the substitution identity sin (-x) = sin x, are perhaps easier to use when solving sin x = k or

cos x = k in radians.).

the solutions of cos x = 0.75 for 0 xare 0.723 c and 5.560

c.




Solution of equations of the form tan x = k is easier: we just add or subtract multiples of 180° (or

multiples of if using radians) to the principal value.

Example (5): Solve tan x = 1.5 for - x, giving the answer in radians to 3 decimal places .

The principal value is 0.983c, but because the tangent function repeats itself every radians, there is

another solution at (0.983 - )c or –2.159

c.


Note that the negative angle is denoted by a clockwise arrow.



Summary. (The principal solution is the one displayed on the calculator.)

The rules are designed to find solutions in the range -180° x < 180° or - x <

This range has the virtue of having the principal solution coincide with the calculator display.

If a different range is stipulated, it is only a matter of adding multiples of 360° or 2 to the solutions so

obtained (when solving equations of the form sin x = k or cos x = k), or multiples of 180° or (when

solving equations of the form tan x = k). In each case, n is an integer.

Solving sin x = k.

Value of k Principal soln. Companion solution Additional solutions

0 x = 0° or 0 rad (none) Add/subtract 180n ° or n

1 x = 90° or /2 (none) Add/subtract 360n ° or 2n

-1 x = -90° or -/2 (none) Add/subtract 360n ° or 2n

positive 0° < x < 90° or 0 < x </2 180° - x or - x Add/subtract 360n ° or 2n

negative -90° < x < 0° or -/2 < x < -180° - x or -- x Add/subtract 360n ° or 2n

Solving cos x = k.


0 x = 90° or /2 (none) Add/subtract 180n ° or n

1 x = 0° or 0 rad (none) Add/subtract 360n ° or 2n

-1 x = -180° or - (none) Add/subtract 360n ° or 2n

positive 0° < x < 90° or 0 < x </2 - x Add/subtract 360n ° or 2n

negative 90° < x < 180° or /2 < x < - x Add/subtract 360n ° or 2n

Solving tan x = k.


0 x = 0° or 0 rad (none) Add/subtract 180n ° or n

positive 0° < x < 90° or 0 < x </2 (none) Add/subtract 180n ° or n

negative -90° < x < 0° or -/2 < x < (none) Add/subtract 180n ° or n



Sometimes we need to use the idea of transformations to solve slightly more complex trig equations.

Example (6): Use the results from Example (4) to solve cos 2x = 0.75 for 0 x, giving your

answers in radians to three decimal places.

Here the limits for x are 0 x, but we are asked to solve for 2x. To ensure that no values are

omitted, we must substitute A for 2x and multiply the limiting values by 2 to get the transformed limit

of 0 A2

From Example (4), we see that the solutions of cos A = 0.75 for 0 Aare 0.723 c and 5.560

c.

To convert the A-values back to x-values, we must divide by 2.

the solutions of cos 2x = 0.75 for 0 xare 0.361 c and 2.780

c.

Example (7): Solve tan(2x + 60°) = 1 for 0 x360°.

By letting A stand for 2x + 60°, we first transform the x-limits to A-limits:

0 x360° 60° A° (x-values doubled, 60° added).

When drawing sketch graphs or CAST

diagrams, draw them with respect to the

transformed variable A, and not the

original variable x.

The substitution back to x-values must

be done afterwards.

The principal value of A where tan A =

1 is 45°, but that value is not within the

A-limits.

We therefore keep adding multiples of

180° to obtain A = 225°, 405°, 585° and

765°.

Since we doubled the x-values and then added 60° to get the A-values, we must perform the inverse

operations to change the A-values back to x-values – i.e. subtract 60° and then halve the result.

Therefore A = 225° x = ½(225-60)° = 82.5°.

Similarly A = 405° x = 172.5°, A = 585° x = 262.5° and finally A = 765° x = 352.5°.

The solutions in the range are therefore x = 82.5°, 172.5°, 262.5° and 352.5°.



Example (8): Solve sin (3x - 40)° = -0.3 for 0 x0° giving answers in degrees to one decimal

place.

Again, we must substitute the x-limits of 0 x0° with A – limits. By using A = 3x- 40, the

A-limits transform to –40° A 00°.

The graph below uses the A-limits.

The principal value of A is here –17.5°,. which is inside the A-limits.

This value is 107.5° to the left of the line of symmetry at A= 90°, so another solution will be 107.5° to

the right, i.e. at A = 197.5°

(We could also have used sin (180-x)° = sin x°, and 180° - (-17.5°) = 197.5°.)

CAST diagram working:

Note that the negative

angle is labelled with a

clockwise arrow.

To complete the full set of solutions for A in the required range, we add multiples of 360° to the two

above values. In fact the only additional one is A = (-17.5 + 360)° = 342.5°.

These three solutions will then need converting back from A-values to x-values by the inverse operation

of

3

40

Ax , so when A = -17.5°, x = 7.5°.

Similarly, when A = 197.5°, x = 79.2° and when A = 342.5°, x = 127.5°.

Hence the solutions of sin (3x - 40)° = -0.3 for 0 x0° are 7.5°, 79.2° and 127.5°.



Quadratic equations in the trigonometric functions are solved in the same way, but care must be taken

when factorising them.

Note also that the graphs accompanying the solutions are for illustration only – you will not be asked to

sketch them !

Example (9) : Solve the equation 2cos2 x - cos x = 0 for - x.

This factorises at once into (cos x) (2 cos x –1) = 0.

cos x = 0 when x = 2 or x = -2.

2 cos x = 1, or cos x = 0.5, when x = 3 or x = -3.

The solutions to the equation are therefore x = 2 and 3 (illustrated below).

Important: we cannot simply cancel out cos x from the equation as follows:

2cos2 x - cos x = 0 2cos

2 x = cos x 2cos x = 1 x = 3.

The final division of both sides by cos x has led to a loss of the solutions satisfying cos x = 0,

i.e. x = 2.



Example (10) : Solve the equation cos2 x - cos x - 1 = 0 for 0 x2. Give the answer in radians to

three decimal places.

This quadratic does not factorise and so the general formula must be used.

Substitute x = cos x, a = 1, b = -1 and c = -1 into the formula.

a

acbbx

2

42

2

51cos

x .

These are the possible solutions, and are 1.618 and -0.618 to 3 decimal places.

The first value can be rejected, since the cosine function cannot take values outside the range

–1 cos x 1.

The only solutions are those where cos x = 2

51. The principal value of x is 2.237

c .

Since cos (2-x) = cos x, another solution would be (6.283–2.237)c or 4.046

c.

the solutions of cos2 x - cos x - 1 = 0 where 0 x2are 2.237

c and 4.046

c to 3 decimal places.



Example (11): Find the angle(s) between 0° and 360° satisfying the equation

1 - 2cos2 x + sin x = 0

This equation can be manipulated into a quadratic in sin x by replacing 2cos2 x with 2(1 -sin

2 x)

using the Pythagorean identity.

1 – 2(1-sin2 x) + sin x = 0

1 – 2 + 2sin2 x + sin x = 0

2sin2 x + sin x – 1 = 0

This is now a quadratic in sin x which factorises into (2 sin x - 1)(sin x + 1) = 0

Hence sin x = 0.5 or –1.

This gives x = 30° or 150° (where sin x = 0.5); also x = 270° (where sin x = -1).

Example (12): Prove that xxx

xxtan

cossin

)cos1)(cos1(

.

The top line of the left-hand expression can be recognised as a ‘difference of squares’ result,

namely 1-cos2 x. This in turn can be replaced by sin

2 x using the Pythagorean identity.

The expression on the left thus becomes xx

x

cossin

sin 2

, and dividing both sides by sin x we finally

obtain x

x

cos

sin or tan x.



Example (13): Show that the equation sin x - cos2 x – 5 = 0 has no solutions.

Substituting cos2 x by 1 -sin

2 x gives sin x – (1-sin

2 x) – 5 = 0

sin x – 1 + sin2 x –5 = 0

sin2 x + sin x – 6 = 0

sin x + 3) (sin x – 2) = 0

sin x = 2 or –3.

The quadratic in sin x factorises all right, but the solutions are impossible because the sine function

cannot take values outside the range –1 sin x 1.

Example (14): Solve the equation sin x = xcos3 for angles between 0° and 360°.

We can divide the equation by cos x to give 3cos

sin

x

x, or 3tan x .

The principal value of x is 60°, but since the tangent graph repeats every 180°, another solution is x =

240°.

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